NCEA Level 2 Physics (91173) 2014 — page 1 of 6
Assessment Schedule – 2014
Physics: Demonstrate understanding of electricity and electromagnetism (91173)
For questions 1,2,4 :
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response; no
relevant evidence.
1a
2a
3a
4a
2m + 1a
2m + 2a
2m + 1e
1e + 2m + 1a
Other combinations are also possible. However, for M5 or M6 at least one Merit question needs to be correct. For E7 or E8 at
least one Excellence needs to be correct.
For question 3 :
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response; no
relevant evidence.
Relevant
evidence
1a
2a
3a
1m+2a
2m + 1a
1e+1m+1a
2m + 1e
Other combinations are also possible. However, for M5 or M6 at least one Merit question needs to be correct. For E7 or E8 at
least one Excellence needs to be correct.
Evidence Statement
Q
ONE
(a)
(b)
Evidence
Achieved
Electric force upwards.
Gravitational force downwards.
Electric force upwards.
Gravitational force
downwards.
Fel = Eq = 610 × 24 × 10–10
Correct answer to electric
force.
Correct answer for
mass.
1.464 × 10–6
1.49 × 10–7
The oil drop would
accelerate (move) towards
the positive plate.
The field strength
would increase when
distance between plates
is decreased, so the
force on the oil drop
would increase.
–6
= 1.464 × 10 N
Electric force upwards = mg
m=
(c)
Merit
1.464 × 10−6
= 1.49 × 10−7 kg
9.8
The oil drop would accelerate
(move) towards the positive plate.
The field strength would increase
when distance between plates is
decreased, so the force on the oil
drop would increase.
Excellence
NCEA Level 2 Physics (91173) 2014 — page 2 of 6
(d)
They would both experience the
same electric force, since the
electric field strength is the same,
and both have the same sized
charge on them.
The electron would experience a
greater acceleration, since it is
much lighter (has smaller mass)
than the proton.
They would both
experience the same
electric force.
OR
The electron would
experience a greater
acceleration.
They would both
experience the same
electric force, since the
electric field strength is
the same, and both have
the same sized charge
on them.
OR
The electron would
experience a greater
acceleration, since it is
much lighter (has
smaller mass) than the
proton.
They would both
experience the
same electric
force, since the
electric field
strength is the
same, and both
have the same
sized charge on
them.
AND
The electron
would experience
a greater
acceleration, since
it is much lighter
(has smaller mass)
than the proton.
NCEA Level 2 Physics (91173) 2014 — page 3 of 6
TWO
(a)
Rparallel
⎛ 1
1 ⎞
=⎜
+
⎝ 6.8 3.5 ⎟⎠
−1
Rparallel = 2.31 Ω
Recognition that the 3 Ω
resistor is in series with a
parallel combination.
RT = 3.0 + 2.3 = 5.3 Ω
(b)
Voltage across battery V = IR
= 6.0 × 5.3 = 31.8 V
Voltage across 3.5 Ω resistor
= 31.86 – 18 = 13.8 V
=14 V
−1
Rparallel = 2.3 Ω
RT = 3.0 + 2.3 = 5.3 Ω
Voltage across battery V =
IR
= 6.0 × 5.3 = 31.8 V
Voltage across 3.5 Ω
resistor
= 31.8 – 18 = 14 V
Or consistency from part
(a).
(c)
2sf. The final answer cannot be
any more accurate than the least
number of significant figures in
the question.
2sf. The final answer
cannot be any more
accurate than the least
number of sf in the
question.
(d)
(The energy per second is its
power output.)
Power depends on voltage and
current.
Since both resistors are on the
same branch, they both have the
same current through them.
Since voltage is directly
proportional to resistance, when
current is the same (V = IR), the
4.6 Ω resistor would have a
greater voltage across it.
• (The energy per second
is its power output.)
• Power depends on
voltage and current.
Hence the 4.6 Ω resistor would
use greater energy per second.
Rparallel
⎛ 1
1 ⎞
=⎜
+
⎝ 6.8 3.5 ⎟⎠
Since both resistors are
on the same branch,
they both have the same
current through them.
OR
Since voltage is directly
proportional to
resistance, when current
is the same (V = IR), the
4.6 Ω resistor would
have a greater voltage
across it.
• (The energy per
second is its
power output.)
• Power depends
on voltage and
current.
• Since both
resistors are on
the same branch,
they both have
the same current
through them.
• Since voltage is
directly
proportional to
resistance, when
current is the
same (V = IR),
the 4.6 Ω
resistor would
have a greater
voltage across it.
• Hence the 4.6 Ω
resistor would
use greater
energy per
second.
NCEA Level 2 Physics (91173) 2014 — page 4 of 6
THREE
(a)
Lamp A would be dimmer.
Lamp A would be dimmer.
(b)
Adding another lamp in parallel
will mean effective resistance
decreases.
This means the circuit current
will increase.
Hence voltage across lamp A will
increase.
The voltage across lamps B, C,
and D will decrease.
Effective resistance
decreases.
Voltage across B / C / D
will decrease.
V across A increase
(c)
V = IR
V = IR
V = 0.5 × 6.0 = 3.0 V
V = 0.5 × 6.0 = 3.0 V
Voltage across each branch
= 12 – 3 = 9.0 V.
Current through bottom branch
= V / R = 9.0 / 6.0 = 1.5 A
Current through middle branch
= 6.0 – 1.5 = 4.5 A
Resistance of middle branch
= V / I = 9.0 / 4.5 = 2.0 Ω
Voltage across each branch
= 12 – 3 = 9.0 V.
(d)
Resistance of resistor = 2.0 Ω
Adding another lamp in
parallel will mean
effective resistance
decreases.
This means the circuit
current will increase.
Adding another
lamp in parallel
will mean effective
resistance
decreases.
This means the
circuit current will
increase.
Hence voltage
across lamp A will
increase.
The voltage across
lamps B, C, and D
will decrease.
Current through bottom
branch
= V / R = 9.0 / 6.0
= 1.5 A
Current through middle
branch
= 6.0 – 1.5 = 4.5 A
Resistance of
middle branch
= V / I = 9.0 / 4.5
= 2.0 Ω
Resistance of
resistor = 2.0 Ω
NCEA Level 2 Physics (91173) 2014 — page 5 of 6
F = Bqv
= 0.65 × 1.6 × 10–19 × 4.8 × 103
= 4.992 × 10–16 N
F = Bqv →
0.65 × 1.6 × 10–19 × 4.8 ×
103
= 4.992 × 10–16 N
(b)
Conventional current will be
anticlockwise (up the rod).
Conventional current will
be anticlockwise (up the
rod).
(c)
• The magnetic force on the
electrons causes charge
separation.
• Charge separation results in the
formation of an electric field.
• This results in an induced
voltage across the ends of the
wire.
• Since it is a complete circuit,
there will be a current.
OR
(Alternate answer)
• The motion of the rod causes
electrons in the rod to move to
the bottom of the rod.
• Top of rod is positive and
bottom of rod is negative.
• This means there is a voltage.
• This causes electrons to flow
around the loop (or causes a
current to flow around the
loop).
• The magnetic force on
the electrons causes
charge separation.
• Charge separation results
in the formation of an
electric field.
OR
(Alternate answer)
• The motion of the rod
causes electrons in the
rod to move to the
bottom of the rod.
FOUR
(a)
• The magnetic force
on the electrons
causes charge
separation.
• Charge separation
results in the
formation of an
electric field.
• This results in an
induced voltage
across the ends of the
rod.
OR
(Alternate answer)
• The motion of the rod
causes electrons in
the rod to move to the
bottom of the rod.
• Top of rod is positive
and bottom of rod is
negative.
• This means there is a
voltage.
• The magnetic
force on the
electrons causes
charge
separation.
• Charge
separation
results in the
formation of an
electric field.
• This results in an
induced voltage
across the ends
of the wire.
• Since it is a
complete circuit,
there will be a
current.
OR
(Alternate answer)
• The motion of
the rod causes
electrons in the
rod to move to
the bottom of the
rod.
• Top of rod is
positive and
bottom of rod is
negative.
• This means there
is a voltage.
This causes
electrons to flow
around the loop (or
causes a current to
flow around the
loop).
NCEA Level 2 Physics (91173) 2014 — page 6 of 6
(d)
V = BvL
V = 0.85 × 0.40 × 4.6
= 1.564 V
V = BvL →
V = 0.85 × 0.40 × 4.6
= 1.564 V
V
1.564
→
R
0.68
I = 2.3 A
I=
V 1.564
=
R 0.68
I = 2.3 A
F = BIL
I=
F = BIL
F=
0.85 × 2.3 × 0.40
= 0.782 N
F = 0.78 N
F = 0.85 × 2.3 × 0.40 = 0.782 N
F = 0.78 N
Cut Scores
Score range
Not Achieved
Achievement
Achievement with
Merit
Achievement with
Excellence
0–8
9 – 16
17 – 24
25 – 32